The set of equations

                                       x+y+z=1

                                       ax-ay +3z =5

                                       5x—3y+az =6

has infinite solutions, if a =

The correct option is 4.

To find the value of $a$ for which the given system of linear equations has infinitely many solutions, we can write the system in matrix form $AX=B$, where:
$A=\left[\begin{array}{ccc}1& 1& 1\\ a& -a& 3\\ 5& -3& a\end{array}\right]$
is the coefficient matrix, and
$B=\left[\begin{array}{c}1\\ 5\\ 6\end{array}\right]$
is the constant matrix.

For a system of three linear equations in three variables to have infinite solutions, the determinant of the coefficient matrix, denoted as $D$ or $\left|A\right|$, must be equal to zero. Let us compute this determinant:
$D=\left|\begin{array}{ccc}1& 1& 1\\ a& -a& 3\\ 5& -3& a\end{array}\right|$

Expanding the determinant along the first row:

Simplifying each term inside the brackets:
$D=1\times [-a^2+9]-1\times [a^2-15]+1\times [-3a+5a]$
$D=-a^2+9-a^2+15+2a$
$D=-2a^2+2a+24$

Setting the determinant $D=0$:
$-2a^2+2a+24=0$

Dividing the entire equation by $-2$:
$a^2-a-12=0$

Factoring the quadratic equation:
$(a-4)(a+3)=0$
This gives two potential values for $a$:
$a=4$ or $a=-3$.

For infinite solutions, the system must also be consistent. Let's test the given correct value $a=4$ using the equations:
1) $x+y+z=1$
2) $4x-4y+3z=5$
3) $5x-3y+4z=6$

If we add equation (1) and equation (2):
$(x+y+z)+(4-4y+3z)=1+5$
$5x-3y+4z=6$
This matches equation (3) exactly. Since equation (3) is a direct linear combination of equations (1) and (2), the equations are consistent and dependent, confirming that the system has infinitely many solutions when $a=4$.